Ammonium carbamate solid \(\left(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\right)\) decomposes at \(313 \mathrm{~K}\) into ammonia and carbon dioxide gases. At equilibrium, analysis shows that there are \(0.0451\) atm of \(\mathrm{CO}_{2}, 0.0961\) atm of ammonia, and \(0.159 \mathrm{~g}\) of ammonium carbamate. (a) Write a balanced equation for the decomposition of one mole of \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\) (b) Calculate \(K\) at \(313 \mathrm{~K}\).

Short Answer

Expert verified
(a) The balanced chemical equation for the decomposition of ammonium carbamate is: NH₄CO₂NH₂(s) ⇌ 2 NH₃(g) + CO₂(g) (b) The equilibrium constant (K) at 313 K for the decomposition of ammonium carbamate is 0.0267.

Step by step solution

01

Write a balanced chemical equation

For the decomposition of ammonium carbamate, we can write the balanced chemical equation as follows: NH\(_4\)CO\(_2\)NH\(_2\)(s) \(\rightleftharpoons\) 2 NH\(_3\)(g) + CO\(_2\)(g)
02

Calculate the moles of ammonium carbamate at equilibrium

Given that there are 0.159 g of ammonium carbamate at equilibrium, we can determine the number of moles of the compound. The molar mass of NH\(_4\)CO\(_2\)NH\(_2\) is: (1 + 4 + 12 + 16 + 14) g/mol = 60 g/mol To find the moles, divide the mass by the molar mass: moles of NH\(_4\)CO\(_2\)NH\(_2\) = \(\frac{0.159 \, g}{60 \, g/mol}\) = 0.00265 mol Calculate the total moles present at equilibrium: Total moles = moles of NH\(_4\)CO\(_2\)NH\(_2\) + moles of NH\(_3\) + moles of CO\(_2\) Total moles = 0.00265 moles
03

Calculate the equilibrium partial pressures of NH\(_3\) and CO\(_2\)

We are given the partial pressures in atm: P(NH\(_3\)) = 0.0961 atm P(CO\(_2\)) = 0.0451 atm
04

Calculate the equilibrium concentrations of NH\(_3\) and CO\(_2\)

Use the ideal gas law (PV = nRT) to convert the partial pressures to concentrations in moles/L. We have pressure (P) in atm and temperature (T) in K, the gas constant (R) in L.atm / (K.mol), R = 0.0821 L.atm / (K.mol) and T = 313 K Rearrange the ideal gas law to solve for concentration, C = n/V: C = \(\frac{P}{RT}\) Calculate the equilibrium concentrations for ammonia and carbon dioxide: C(NH\(_3\)) = \(\frac{0.0961 \, atm}{0.0821 \, L . atm / (K . mol) \times 313 \, K}\) = 0.0385 M C(CO\(_2\)) = \(\frac{0.0451 \, atm}{0.0821 \, L . atm / (K . mol) \times 313 \, K}\) = 0.0178 M
05

Calculate the equilibrium constant \(K\)

The expression for the equilibrium constant for this reaction is: \(K = \frac{[NH_3]^2 \cdot [CO_2]}{1}\) (since the concentration of the solid NH\(_4\)CO\(_2\)NH\(_2\) does not affect the equilibrium constant) Plug in the equilibrium concentrations to find K: \(K = \frac{(0.0385 \, M)^2 \cdot 0.0178 \, M}{1}\) = 0.0267 So the equilibrium constant K at 313 K is 0.0267.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At \(460^{\circ} \mathrm{C}\), the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g)$$ has \(K=84.7\). All gases are at an initial pressure of \(1.25\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

The reaction $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ has an equilibrium constant of \(1.30\) at \(650^{\circ} \mathrm{C}\). Carbon monoxide and steam both have initial partial pressures of \(0.485 \mathrm{~atm}\), while hydrogen and carbon dioxide start with partial pressures of \(0.159\) atm. (a) Calculate the partial pressure of each gas at equilibrium. (b) Compare the total pressure initially with the total pressure at equilibrium. Would that relation be true of all gaseous systems?

Given the following descriptions of reversible reactions, write a balanced equation (simplest whole-number coefficients) and the equilibrium constant expression \((K)\) for each. (a) Nitrogen gas reacts with solid sodium carbonate and solid carbon to produce carbon monoxide gas and solid sodium cyanide. (b) Solid magnesium nitride reacts with water vapor to form magnesium hydroxide solid and ammonia gas. (c) Ammonium ion in aqueous solution reacts with a strong base at \(25^{\circ} \mathrm{C}\), giving aqueous ammonia and water. (c) Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions.

Consider the system $$\mathrm{A}(g)+2 \mathrm{~B}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{D}(g)$$ at \(25^{\circ} \mathrm{C}\). At zero time, only \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are present. The reaction reaches equilibrium \(10 \mathrm{~min}\) after the reaction is initiated. Partial pressures of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{D}\) are written as \(P_{\mathrm{A}}, P_{\mathrm{B}}\), and \(P_{\mathrm{D}}\). Answer the questions below, using LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) \(P_{\mathrm{D}}\) at \(11 \mathrm{~min}\) ________ \(P_{\mathrm{D}}\) at \(12 \mathrm{~min} .\) (b) \(P_{\mathrm{A}}\) at \(5 \mathrm{~min}\) \(P_{\mathrm{A}}\) ______ at \(7 \mathrm{~min}\) (c) \(K\) for the forward reaction ______ \(K\) for the reverse reaction. (d) At equilibrium, \(K\)______Q. (e) After the system is at equilibrium, more of gas \(\mathrm{B}\) is added. After the system returns to equilibrium, \(K\) before the addition of \(B\) \(K\) _____ after the addition of \(\mathrm{B}\). (f) The same reaction is initiated, this time with a catalyst. \(K\) for the system without a catalyst _____ \(K\) for the system with a catalyst. (g) \(K\) for the formation of one mole of \(\mathrm{D}\) \(K\) _____ for the formation of two moles of \(\mathrm{D}\). (h) The temperature of the system is increased to \(35^{\circ} \mathrm{C} . P_{\mathrm{B}}\) at equilibrium at \(25^{\circ} \mathrm{C} \longrightarrow P_{\mathrm{B}}\) _______at equilibrium at \(35^{\circ} \mathrm{C}\). (i) Ten more grams of \(\mathrm{C}\) are added to the system. \(P_{\mathrm{B}}\) before the addition of \(\mathrm{C} \quad P_{\mathrm{B}}\) _____ after the addition of \(\mathrm{C}\).

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{CO}_{2}}\right)^{3}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{4}}{\left(P_{\mathrm{C}, \mathrm{H}_{4}}\right)\left(P_{\mathrm{O}_{2}}\right)^{5}}\) (b) \(K=\frac{P_{\mathrm{C}_{0} \mathrm{H}_{12}}}{\left(P_{\mathrm{C}, \mathrm{H}_{6}}\right)^{2}}\) (c) \(K=\frac{\left[\mathrm{PO}_{4}^{3-}\right]\left[\mathrm{H}^{+}\right]^{3}}{\left[\mathrm{H}_{3} \mathrm{PO}_{4}\right]}\) (d) \(K=\frac{\left(P_{\mathrm{CO}_{2}}\right)\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)}{\left[\mathrm{CO}_{3}^{2-}\right]\left[\mathrm{H}^{+}\right]^{2}}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free